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1 Topology Proceedings Web: Mail: Topology Proceedings Department of Mathematics & Statistics Auburn University, Alabama 36849, USA ISSN: COPYRIGHT c by Topology Proceedings. All rights reserved.

2 TOPOLOGY PROCEEDINGS Volume 28, No. 1, 2004 Pages SYMMETRIC PRODUCTS AS CONES AND PRODUCTS ENRIQUE CASTAÑEDA ALVARADO Abstract. In this paper we prove that: (a) if X is a finite graph, then the second symmetric product F 2 (X) = {{p, q} X : p, q X} of X is the cone over some continuum Y, if and only if X is a simple n-od or an arc, and (b) if X is a finite graph, then F 2 (X) is a product of two nondegenerate continua, if and only if X is an arc. 1. Introduction A continuum is a compact, connected metric space. Let N denote the set of positive integers. Let n N. Consider the following hyperspaces of a continuum X: C(X) = {A X : A is a closed, nonempty connected subset of X}, F n (X) = {A X : A is a nonempty subset of X with at most n elements}. Both hyperspaces are considered with the Hausdorff metric. The problem of determing those finite dimensional continua X for which C(X) is homeomorphic to cone(x) has been extensively studied. (See [1], [4], [8], [9], [10], [11], [12], [14], [15], [17], [20], [21], [22] and [23].) A detailed discussion about this topic can be found in [11, sections 7 and 80]. The case that X is hereditarily decomposable has been completely solved by S. B. Nadler, Jr. in [17] Mathematics Subject Classification. Primary 54B20. Key words and phrases. arc, cone, continuum, finite graph, hyperspace, simple n-od, symmetric product. 55

3 56 E. CASTAÑEDA ALVARADO He showed that there exist exactly eight such continua, pictured on page 63 of [11]. In the case that X contains an indecomposable subcontinuum Y, it is known that Y is unique [11, Theorem 80.12], X Y is arcwise connected [8], and Y has the cone = hyperspace property. A characterization in terms of selections of finite dimensional continua Y with the cone = hyperspace has recently appeared in [9]. Continua X for which there exists a finite dimensional continuum Z such that C(X) is homeomorphic to cone(z) have been completely described. Using a previous result by S. Macías [15], A. Illanes and M. de J. López [10] give a complete list of that continua X, in the case that X is hereditarily decomposable, and López [14] gives a precise description of that continua X, in the case that X is not hereditarily decomposable. With respect to products, Illanes [11, Theorem 79.2] has shown that a continuum X has the properties that C(X) is finite dimensional and it is homeomorphic to the product of two nondegenerate continua if and only if X is an arc or a simple closed curve. The hyperspace F n (X) is the so called n th symmetric product of X. Symmetric products were introduced by K. Borsuk and S. Ulam in [2]. They proved that, for I = [0, 1] and n = 1, 2, 3, F n (I) is homeomorphic to I n. It is known that F 2 (S 1 ) is homeomorphic to Möebius strip (see [2, p. 877] or [11, 1.26]). Given an n-od T n (see the definition below in conventions), it was proved in [3, Lemma 1] that F 2 (T n ) is homeomorphic to the cone over a continuum Z. In this paper we consider the problem of determining the finite graphs X such that F 2 (X) is homeomorphic to a cone or to the product of two nondegenerate continua. We prove that: Theorem If X is a finite graph, then F 2 (X) is homeomorphic to the product of two nondegenerate continua if and only if X is an arc. Theorem If X is a finite graph, then F 2 (X) is homeomorphic to the cone over a continuum Z if and only if X is a simple n-od or an arc. 2. Conventions A finite (connected) graph is a continuum which is a finite union of arcs (called edges) such that every two of them meet at a subset

4 SYMMETRIC PRODUCTS AS CONES AND PRODUCTS 57 of their end points. If X is a finite graph, in X are defined edges and vertices. The vertices of X are the end points of the edges of X. A finite graph which is different from a simple closed curve is called an acircular graph. We are interested in distinguishing the ramification points of the graph X from the rest of the points, so we assume the each vertex of an acircular graph X is either an end point of X or a ramification point of X. With this restriction, the two end points of an edge of X may coincide and such an edge is a simple closed curve. These edges will be called loops. Thus, the edges of X are arcs or simple closed curves, and in X there are only three kinds of edges: loops, edges that contain some end point, and edges joining ramification points. We assume that the metric d in X is the metric of arc length and each edge of X has length equal to one. The set of ramification points of X is denoted by R(X). A simple n-od T n is a finite graph which is the union of n arcs J 1,..., J n such that there exists a point p T n with the property J i J j = {p}, if i j, and p is an end point of each one of the arcs J i. The point p is called the core of T n. A simple 3-od is called a simple triod. Given n, m N, where n, m 3, let K n,m denote the complete bipartite graph; that is, K n,m contains n + m vertices {v 1,..., v n, w 1,..., w m } and n m edges v i w j, where v i w j joins v i and w j, for every i {1,..., n} and j {1,..., m}. Given a continuum Z and a subset A of Z, bd Z (A), cl Z (A), and int Z (A) denote the respective boundary, closure, and interior of A in Z. Let Z be a continuum and p Z. Let β be a cardinal number. We say that p is of order less than or equal to β in Z, written ord(p, Z) β, provided that for each open subset U of Z such that p U, there exists an open subset V of Z such that p V U and bd Z (V ) β. We say the p is of order β, written ord(p, Z) = β, provided that ord(p, Z) β and ord(p, Z) α for any cardinal number α < β. A point p Z is called an end point of Z provided that ord(p, Z) = 1. A point p Z is called a ramification point of Z provided that ord(p, Z) 3. If A is a subset of Z, p Z, and ε > 0, let B Z (ε, p) = {q Z : d Z (p, q) < ε} and N Z (ε, A) = {q Z : there exists p A such that d Z (p, q) < ε}. A Peano continuum is a locally connected continuum. Given open subsets U 1,..., U m of a continuum X, let U 1,..., U m n = {A F n (X) : A U 1... U m

5 58 E. CASTAÑEDA ALVARADO and A U i }. It is known that the sets of the form U 1,..., U m n form a basis for the topology of F n (X). 3. Symmetric products as cones or products Lemma 3.1. Let T n and T m denote a simple n-od and a simple m- od, respectively. If 2 m < n, then T n [0, 1] cannot be embedded in T m [0, 1]. Proof: Let X = T n [0, 1] and Y = T m [0, 1]; let T n = pa 1... pa n and T m = qb 1... qb m, where the sets pa i and qb j are arcs, pa i pa j = {p}, if i j and qb k qb r = {q}, if k r. Let C X = {p} [0, 1], C X = {p} (0, 1), C Y = {q} [0, 1], and C Y = {q} (0, 1). Suppose that there exists an embedding h : X Y. Claim. h(c X ) C Y. Suppose to the contrary that there exists a point (p, t) C X such that h(p, t) = (y, s) and y q. We may assume that y qb 1 {q}. Then there exists a simple n-od T T n and an arc J [0, 1] such that (p, t) T J and h(t J) (qb 1 {q}) [0, 1]. This implies that the product T J can be embedded in [0, 1] 2. By using the Theorem on the Invariance of the Domain [6, Theorem VI 9, p. 95], it can be shown that this is impossible. This contradiction completes the proof of the claim. Fix a point x 0 C X. Since h(c X ) is homeomorphic to R and it is contained in the arc C Y, we have that C Y h(c X ) is compact and x 0 / h 1 (C Y h(c X )). Thus, there exists a simple n-od T T n and an arc J (0, 1) such that x 0 T J X h 1 (C Y h(c X )). Let D = {p} J. Given a point x (T J) D, h(x) / C Y h(c X ). If h(x) C Y, then h(x) h(c X ). Thus, x C X (T J) = D. This contradicts the choice of x and proves that h(x) / C Y. We have shown that h((t J) D) Y C Y. Suppose that T = pc 1... pc n, where each c i pa i {p}. Since the components of h((t J) D) are the sets h((pc 1 {p}) J),..., h((pc n {p}) J) and the components of Y C Y are the separated sets (qb 1 {q}) [0, 1],..., (qb m {q}) [0, 1] and m < n, we may assume that h((pc 1 {p}) J) h((pc 2 {p}) J) (qb 1 {q}) [0, 1] and h((pc 3 {p}) J) ((qb 1 {q}) (qb 2 {q})) [0, 1]. Thus, h(((pc 1 {p}) (pc 2 {p}) (pc 3 {p})) J) (qb 1 qb 2 ) [0, 1].

6 SYMMETRIC PRODUCTS AS CONES AND PRODUCTS 59 Therefore, h((pc 1 pc 2 pc 3 ) J) (qb 1 qb 2 ) [0, 1]. This is impossible since, by using the Theorem on the Invariance of the Domain [6, Theorem VI 9, p. 95], it can be shown that the product of a simple triod and an arc cannot be embedded in the plane. The proof of the lemma is complete. Lemma 3.2. Let T n and T m denote a simple n-od and a simple m-od, respectively, where n, m N and n, m 3. Then T n T m is homeomorphic to the cone over the complete bipartite graph K n,m. Proof: Suppose that T n = n i=1 θe i and T m = m j=1 θ e j, where {e 1,..., e n } and {e 1,..., e m} are the respective canonical basis of the linear spaces R n and R m and {e 1,..., e n } {e 1,..., e m} = ; moreover, θ and θ are the respective origins of R n and R m, and θe i and θ e j are the convex segments joining θ and e i and θ and e j, respectively. Let Z = {(x, y) T n T m : x {e 1,..., e n } or y {e 1,..., e m}}. Claim. Z is a complete bipartite graph K n,m. In order to prove this claim, we need to define the vertices and the edges of Z. Let U = {(e i, θ ) : i {1,..., n}} and V = {(θ, e j ) : j {1,..., m}}. For each i {1,..., n} and each j {1,..., m}, let A i,j = ({e i } θ e j ) (θe i {e j }). Clearly, A i,j is an arc that joins (e i, θ ) and (θ, e j ). Now suppose that there exists a point (x, y) A i,j A k,r and (i, j) (k, r). It is easy to show that (x, y) is an end point of A i,j and of A k,r. We have shown that the set of vertices U V and the set of edges A i,j form a complete bipartite graph K n,m. Since Z = {A i,j : i {1,..., n} and j {1,..., m}}, the claim is proved. Now, let F : cone(z) T n T m be given by F ((x, y), t) = ((1 t)x, (1 t)y). Clearly, F is continuous. In order to show that F is one-to-one, suppose that F ((x, y), t) = F ((u, v), s). Then (1 t)x = (1 s)u and (1 t)y = (1 s)v. By the definition of Z, we may assume that x = e i for some i {1,..., n}. We consider four cases: Case 1: u = e k for some k {1,..., n} and t 1. Since (1 t)e i = (1 s)e k, it follows that t = s and e i = e k. Since (1 t)y = (1 s)v, we conclude that y = v.

7 60 E. CASTAÑEDA ALVARADO Case 2: u = e k for some k {1,..., n} and t = 1. Since (1 t)e i = (1 s)e k, it follows that s = 1. Thus, ((x, y), t) and ((u, v), s) are the same point in cone(z). Case 3: v = e r for some r {1,..., m} and t 1. Since (1 t)e i = (1 s)u, (1 t)e i = (1 s)u. This implies that 1 t 1 s. On the other hand, (1 t)y = (1 s)e r implies that 1 t 1 s. Thus, s = t. Hence, u = e i = x and y = e r = v. Case 4: v = e r for some r {1,..., m} and t = 1. Thus, θ = (1 t)y = (1 s)e r. This implies that s = 1. Thus, ((x, y), t) and ((u, v), s) are the same point in cone(z). Now we show that F is an onto map. Let (x, y) T n T m. Then x = ae i and y = be j for some a, b [0, 1], i {1,..., n}, and j {1,..., m}. We may assume that a b. If b = 0, then (x, y) = (x, θ) = F ((e i, θ), (1 a)). If b 0, then (x, y) = F (((a/b)e i, e j ), 1 b). This completes the proof that F is onto. Therefore, F is a homeomorphism and the proof of the lemma is complete. Given topological spaces X and Y and points p X and q Y, we write (X, p) (Y, q) if there exists a homeomorphism f : X Y such that f(p) = q. Given an n-od T n with core z and end points z 1,..., z n, let Z n = {A F 2 (T n ) : z i A for some i {1,..., n}}. It was proved in [3, Lemma 1] that F 2 (T n ) is homeomorphic to cone(z n ). Lemma 3.3. Let X be a finite graph and p, q X. Then the element A = {p, q} has a basis of neighborhoods B in F 2 (X) with the property that for each U B: (a) if ord(p, X) = 1 or ord(p, X) = 2 and ord(q, X) = 1, then (U, A) ([0, 1] 2, (0, 0)); (b) if ord(p, X) = 2, p q and ord(q, X) = 2, then (U, A) ([0, 1] 2, ( 1 2, 1 2 )); (c) if ord(p, X) = 2 and p = q, then (U, A) ([0, 1] 2, (0, 0)); (d) if ord(p, X) = 1 and ord(q, X) = m 3, then (U, A) ([0, 1] T m, (0, z)), where T m is a simple m-od and z is the core of T m ; (e) if ord(p, X) = 2 and ord(q, X) = m 3, then (U, A) ([0, 1] T m, ( 1 2, z)), where T m is a simple m-od and z is the core of T m ; (f) if ord(p, X) = n 3, p q and ord(q, X) = m 3, then (U, A) (cone(k n,m ), v), where v is the vertex of cone(k n,m );

8 SYMMETRIC PRODUCTS AS CONES AND PRODUCTS 61 (g) if ord(p, X) = n 3, and p = q, then (U, A) (cone(z n ), v), where v is the vertex of cone(z n ), where Z n is as described above. Proof: Given points p, q X such that p q, there exist bases of closed connected neighborhoods K and L of p and q, respectively, in X such that K satisfies the following: (i) if ord(p, X) = 1, then K is an arc and p is an end point of K; (ii) if ord(p, X) = 2, then K is an arc and p is not an end point of K; (iii) if ord(p, X) = n 3, then K is an n-od and p is the core of K. The set L satisfies analogous properties, depending on ord(q, X). Notice that K L is homeomorphic to the neighborhood K, L 2 of A = {p, q} in F 2 (X) (with the homeomorphism that sends the pair (x, y) into the set {x, y}). In the case that p = q, we can select a basis of neighborhoods K of p in X as in the previous paragraph and, in this case, a K 2 = F 2 (K) is a neighborhood of A = {p}. With the observations contained in the previous paragraphs and Lemma 3.2, the proof of this lemma is easy. The following lemma provides a description of the neighborhoods of a point in the product of two finite graphs. The proof is immediate ((e) follows from Lemma 3.2). Lemma 3.4. Let Y and Z be finite graphs y Y and z Z. Then the element A = (y, z) has a basis of neighborhoods B with the property that for each U B: (a) if ord(y, Y ) = 1 or ord(y, Y ) = 2 and ord(z, Z) = 1, then (U, A) ([0, 1] 2, (0, 0)); (b) if ord(y, Y ) = 2 and ord(z, Z) = 2, then (U, A) ([0, 1] 2, ( 1 2, 1 2 )); (c) if ord(y, Y ) = 1 and ord(z, Z) = m 3, then (U, A) ([0, 1] T m, (0, z)), where T m is a simple m-od and z is the core of T m ; (d) if ord(y, Y ) = 2 and ord(z, Z) = m 3, then (U, A) ([0, 1] T m, ( 1 2, z)), where T m is a simple m-od and z is the core of T m ; (e) if ord(p, X) = n 3 and ord(q, X) = m 3, then (U, A) (cone(k n,m ), v), where v is the vertex of cone(k n,m ). Lemma 3.5. Let T 3 and T n denote a simple 3-od and a simple n-od, respectively, with n N and n 3. Then F 2 (T 3 ) cannot be embedded in T n [0, 1]. Proof: Suppose, to the contrary, that there exists an embedding h : F 2 (T 3 ) T n [0, 1]. Let p be the core of T 3 and T = {{p, x}

9 62 E. CASTAÑEDA ALVARADO F 2 (T 3 ) : x T 3 }. Notice that T is homeomorphic to T 3. Let z be the core of T n. Given x T 3 {p}, there exists an arc J in T 3 such that J is a neighborhood of x in T 3 and p / J. Thus, there exists a simple subtriod S of T 3 such that p is the core of S and S J =. Then each neighborhood of {p, x} in F 2 (T 3 ) which is contained in S, J 3 contains a homeomorphic copy of the product S J. Since this space cannot be embedded in the plane, and the components of (T n [0, 1]) ({z} [0, 1]) are planable spaces, we conclude that h({p, x}) {z} [0, 1] for each x p. By continuity, h(t ) {z} [0, 1]. This is impossible since an arc cannot contain a simple triod. This contradiction proves the lemma. Lemma 3.6. Let n, m, r N be such that n 3 and r, m 2. Let T n be a simple n-od with core p. Then there is not a continuum Z, a point z Z and closed neighborhoods A and B of z in Z such that A B, (B, z) (cone(k r,m ),vertex of cone(k r,m )) and (A, z) (F 2 (T n ), {p}). Proof: Suppose, to the contrary, that there exists a continuum Z, a point z Z, and closed neighborhoods A and B of z in Z such that A B, (B, z) (cone(k r,m ),vertex of cone(k r,m )) and (A, z) (F 2 (T n ), {p}). Let h : B cone(k r,m ) and g : F 2 (T n ) A be homeomorphisms such that h(z) = v and g({p}) = z, where v is the vertex of cone(k r,m ). Let W be an open subset of Z such that z W A h 1 (K r,m {0}). Let q T n {p} be such that g({q}) W. Then there exists an arc J T n {p} such that J has end points u and v, J {u, v} is an open subset of T n, q J {u, v}, and g(f 2 (J)) W. Thus, g(f 2 (J {u, v})) is an open subset of A. So, there exists an open subset U of Z such that g(f 2 (J {u, v})) = A U = A U W = U W. Hence, g(f 2 (J {u, v})) is an open subset of Z that contains the point g({q}). Since (F 2 (J {u, v}), {q}) ([0, 1) 2, (0, 0)), (h(g(f 2 (J {u, v}))), h(g({q}))) ([0, 1) 2, (0, 0)). On the other hand, the set V = h(g(f 2 (J {u, v}))) is an open subset of cone(k r,m ) (K r,m {0}) that contains the point y = h(g({q})) and satisfies (V, y) ([0, 1) 2, (0, 0)). However, it is easy to show that K r,m is a finite graph without end points. Thus, it is impossible that the point y cone(k r,m ) (K r,m {0}) has a neighborhood V such that (V, y) ([0, 1) 2, (0, 0)). This contradiction proves the lemma.

10 SYMMETRIC PRODUCTS AS CONES AND PRODUCTS 63 Lemma 3.7. Let X be a finite graph and Y be a Peano continuum. If Y [0, 1] can be embedded in F 2 (X), then ord(y, Y ) is finite for every y Y. Proof: Suppose, to the contrary, that there exists a point y Y such that ord(y, Y ) is infinite and suppose also that there exists an embedding h : Y [0, 1] F 2 (X). Let R = {A F 2 (X) : A R(X)}. Since R is finite, h 1 (R) is finite. Thus, there exists a subarc J of [0, 1] such that h({y} J) F 2 (X) R. Let m = max{ord(x, X) : x X}. Since ord(y, Y ) is infinite, there exists a simple (m + 1)-od T m+1 in Y such that y is the core of T m+1 (See [13, Ch. VI, 51, Example 8, p. 277].) Fix t J. Since h(y, t) F 2 (X) R, h(y, t) is an element of F 2 (X) of one of the forms described in (a), (b), (c), (d) or (e) in Lemma 3.3. Thus, h(y, t) has a neighborhood U in F 2 (X) of the form [0, 1] T, where T is either an arc or an r-od for some r m. By continuity, there exists a subarc L of J and a simple (m + 1)-od S m+1, contained in T m+1 such that h(l S m+1 ) U. Thus, it is possible to embed the product L S m+1 into [0, 1] T. This contradicts Lemma 3.1 and completes the proof of the lemma. Lemma 3.8. For m 3, let T m be a simple m-od and Y a Peano continuum. If Y [0, 1] can be embedded in T m [0, 1], then the number of points y Y such that ord(y, Y ) > 2 is finite. Proof: Let p be the core of T m. Suppose, to the contrary, that the number of points y Y such that ord(y, Y ) > 2 is infinite and there exists an embedding h : Y [0, 1] T m [0, 1]. Choose a sequence of different points y 1, y 2,... such that ord(y n, Y ) > 2 for each n N and lim y n = y 0 for some y 0 Y. Given n N and t [0, 1], each neighborhood of h(y n, t) contains a topological copy of the set T 3 [0, 1], where T 3 is a simple triod (see [13, Ch. VI, 51, Example 8, p. 277] ). Since (Y [0, 1]) ({p} [0, 1]) is a finite union of open sets homeomorphic to the space (0, 1] [0, 1], then h(y n, t) {p} [0, 1]. We have shown that h({y n } [0, 1]) {p} [0, 1] for each n N. By continuity, h({y 0 } [0, 1]) {p} [0, 1]. Thus, the sets h({y 1 } [0, 1]), h({y 2 } [0, 1]),... are pairwise nondegenerate disjoint continua, contained in the arc {p} [0, 1] and they tend to h({y 0 } [0, 1]). Since this situation is impossible for an arc, the lemma is proved.

11 64 E. CASTAÑEDA ALVARADO Lemma 3.9. Let X be a finite graph and Y a Peano continuum. If Y [0, 1] can be embedded in F 2 (X), then the number of points y Y such that ord(y, Y ) > 2 is finite. Proof: Suppose, to the contrary, that the number of points y Y such that ord(y, Y ) > 2 is infinite and there exists an embedding h : Y [0, 1] F 2 (X). Choose a sequence of different points y 1, y 2,... such that ord(y n, Y ) > 2 and lim y n = y 0 for some y 0 Y. Given t [0, 1], each neighborhood of h(y 0, t) contains points of the form h(y n, t), and then these neighborhoods contain topological copies of the set T 3 [0, 1], where T 3 is a simple triod. (See [13, Ch. VI, 51, Example 8, p. 277].) Thus, h(y 0, t) is not of any of the forms described in (a), (b), and (c) of Lemma 3.3. Let A = {A F 2 (X) : A R(X)}. Since A is finite, there exists an arc J [0, 1] such that h({y 0 } J) A =. Therefore, h(y 0, t) is not of any of the forms described in (f) and (g) of Lemma 3.3 for all t [0, 1]. Fix t 0 J. By the previous paragraphs, h(y 0, t 0 ) is of one of the forms (d) or (e) of Lemma 3.3. In both cases, h(y 0, t 0 ) has a neighborhood U of the form T m [0, 1] for some simple m-od T m. Let Y 1 be a closed connected, locally connected neighborhood of y 0 in Y and L a subarc of J such that h(y 1 L) U. Since Y 1 contains infinitely many points y such that ord(y, Y 1 ) > 2, we obtain a contradiction to Lemma 3.8. This finishes the proof of this lemma. Lemma Let X be a finite graph. If F 2 (X) is homeomorphic to the product of two nondegenerate continua Y and Z, then Y and Z are finite graphs. Proof: Since X is locally connected, F 2 (X) is locally connected. (See [2, property (a), p. 877].) Thus, Y and Z are locally connected. Fix an arc J in Z. Then Y J can be embedded in F 2 (X). By lemmas 3.7 and 3.9, ord(y, Y ) is finite for every y Y and the number of points y Y such that ord(y, Y ) > 2 is finite. By [19, Theorem 9.10], Y is a finite graph. Similarly, Z is a finite graph. Theorem Let X be a finite graph. Then F 2 (X) is homeomorphic to the product of two nondegenerate continua Y and Z if and only if X is an arc.

12 SYMMETRIC PRODUCTS AS CONES AND PRODUCTS 65 Proof: Since F 2 ([0, 1]) is homeomorphic to [0, 1] 2, the sufficiency is immediate. Now, we prove the necessity. First, we show that X does not contain ramification points. Suppose, to the contrary, that X contains a ramification point p. Let h : F 2 (X) Y Z be a homeomorphism. By Lemma 3.10, Y and Z are finite graphs. Since the point p has a basis of neighborhoods in X of the form T m, where T m is a simple m-od, p is the core of T m and m = ord(p, X), we have that {p} has a basis of neighborhoods in F 2 (X) of the form F 2 (T m ). By Lemma 3.5, the sets of the form F 2 (T m ) cannot be embedded in sets of the form T n [0, 1], where T n is a simple n-od. By Lemma 3.4, h(p) is of the form h(p) = (y, z) for some y Y and z Z such that ord(y, Y ) = r 3 and ord(z, Z) = s 3. By Lemma 3.4, (y, z) has a basis of neighborhoods B in Y Z with the property that, for each U B, (U, (y, z)) (cone(k r,s ), v), where v is the vertex of cone(k r,s ). Since h is a homeomorphism, (y, z) also has a basis of neighborhoods B 0 in Y Z with the property that, for each V B 0, (V, (y, z)) (F 2 (T m ), {p}). According to Lemma 3.6, this is absurd. Therefore, X does not contain ramification points. Thus X is an arc or a simple closed curve. If X is a simple closed curve, then F 2 (X) is homeomorphic to the Möbius strip. Thus, F 2 (X) is not homeomorphic to the product of two nondegenerate continua. This proves that X is not a simple closed curve. Therefore, X is an arc. Lemma Let X be a finite graph. If F 2 (X) is homeomorphic to the cone over a continuum Y, then Y is a finite graph. Proof: Since X is locally connected, F 2 (X) is also locally connected [2, property (a), p. 877]), so Y is locally connected. Since Y [0, 1 2 ] can be embedded in F 2(X), by lemmas 3.7 and 3.9, ord(y, Y ) is finite for every y Y and the number of points y Y such that ord(y, Y ) > 2 is finite. By [19, Theorem 9.10], Y is a finite graph. Lemma Let X be a finite graph. If F 2 (X) is homeomorphic to the cone over a continuum Y, then X is unicoherent.

13 66 E. CASTAÑEDA ALVARADO Proof: Suppose, to the contrary, that X is not unicoherent. Since X is a locally connected continuum, this implies, by [7, Theorem 1.6], that F 2 (X) is not unicoherent. By [5, theorems 2 and 3] and [16], there exists a continuous function f : F 2 (X) S 1, where S 1 is the unit circle in the plane, such that f is not homotopic to a constant map. Since cone(y ) is contractible, F 2 (X) is contractible and every map from F 2 (X) to S 1 is homotopic to a constant map. This contradiction completes the proof of the lemma. Theorem Let X be a finite graph. Then F 2 (X) is homeomorphic to the cone over a continuum Y if and only if X is a simple n-od or an arc. Proof: (Necessity). By Lemma 3.12, Y is a finite graph. Let h : F 2 (X) cone(y ). First, we show that if p is a ramification point of X and v is the vertex of cone(y ), then h({p}) = v. Suppose, to the contrary, that h({p}) = (y, t) for some t [0, 1). Since Y is a finite graph, (y, t) has a basis of neighborhoods B in cone(y ) such that, for each U B, U is of the form [0, 1] T, where T is an arc of a simple r-od. On the other hand, {p} has a basis of neighborhoods of the form F 2 (T m ), where m = ord(p, X) and T m is a simple m-od. By Lemma 3.5, the basic neighborhoods of {p} cannot be embedded in the basic neighborhoods of (y, t). This contradiction proves that h({p}) = v. Since h is one-to-one, we conclude that X has at most one ramification point. Since X is unicoherent (Lemma 3.13), we conclude that X is an arc or a simple n-od. (Sufficiency). Is immediate from [3, Lemma 1]. Question Is [0, 1] the only finite graph such that F 3 (X) is a product of two nondegenerate continua? By [2, Theorem 6], F 3 ([0, 1]) is homeomorphic to [0, 1] 3. References 1. F. D. Ancel and S. B. Nadler, Jr., Cones that are cells, and an application to hyperspaces, Topology Appl. 98 (1999), K. Borsuk and S. Ulam, On symmetric products of topological spaces, Bull. Amer. Math. Soc. 37 (1931), E. Castañeda, Embedding symmetric products in Euclidean spaces, in Continuum Theory. Lecture Notes in Pure Appl. Math., 230. New York: Dekker,

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